In: Statistics and Probability
How to Run descriptives and a two-tailed, two sample assuming equal variance t-test.
Here's your data:
Weight of Apples in Grams | |||
Apple ID | Farm A | Farm B | |
1 | 131 | 151 | |
2 | 147 | 159 | |
3 | 134 | 162 | |
4 | 134 | 158 | |
5 | 136 | 159 | |
6 | 137 | 160 | |
7 | 140 | 150 | |
8 | 134 | 160 | |
9 | 136 | 160 | |
10 | 133 | 160 | |
11 | 134 | 160 | |
12 | 132 | 158 | |
13 | 139 | 162 | |
14 | 136 | 160 | |
15 | 135 | 154 | |
16 | 135 | 155 | |
17 | 135 | 159 | |
18 | 138 | 151 | |
19 | 134 | 151 | |
20 | 149 | 150 | |
21 | 149 | 155 | |
22 | 135 | 150 | |
23 | 148 | 150 | |
24 | 135 | 158 | |
25 | 149 | 152 | |
26 | 137 | 162 | |
27 | 134 | 158 | |
28 | 140 | 161 | |
29 | 138 | 160 | |
30 | 133 | 161 | |
31 | 140 | 158 | |
32 | 138 | 160 | |
33 | 150 | 149 | |
34 | 135 | 160 | |
35 | 136 | 160 | |
36 | 148 | 157 | |
37 | 150 | 152 | |
38 | 136 | 150 | |
39 | 142 | 157 | |
40 | 132 | 149 | |
41 | 133 | 160 | |
42 | 147 | 150 | |
43 | 135 | 159 | |
44 | 140 | 153 | |
45 | 132 | 151 | |
46 | 132 | 160 | |
47 | 136 | 151 | |
48 | 141 | 162 | |
49 | 136 | 162 | |
50 | 135 | 160 | |
51 | 142 | 160 | |
52 | 135 | 150 | |
53 | 132 | 158 | |
54 | 140 | 160 | |
55 | 136 | 152 | |
56 | 138 | 160 | |
57 | 132 | 159 | |
58 | 136 | 150 | |
59 | 136 | 161 | |
60 | 135 | 158 | |
61 | 135 | 152 | |
62 | 136 | 154 | |
63 | 137 | 151 | |
64 | 138 | 150 | |
65 | 137 | 150 | |
66 | 136 | 153 | |
67 | 134 | 153 |
as the data is large enough i am using MINITAB to solve the problem.
steps:-
copy the data in minitab stat
basic
statistics
2 sample
t
choose each
sample in its own column
in sample 1
select 'farm A' and in sample 2 select 'farm B'
click
options
type 95 in
confidence interval
type 0.0 in
hypothesized difference
choose the
alternative hypothesis difference
hypothesized
difference
tick assume
equal variance
ok
ok.
your output be :-
Two-Sample T-Test and CI: Farm A, Farm B
Method
μ₁: mean of Farm A |
µ₂: mean of Farm B |
Difference: μ₁ - µ₂ |
Equal variances are assumed for this analysis.
Descriptive Statistics
Sample | N | Mean | StDev | SE Mean |
Farm A | 67 | 137.55 | 5.02 | 0.61 |
Farm B | 67 | 156.07 | 4.43 | 0.54 |
Estimation for Difference
Difference | Pooled StDev |
95% CI for Difference |
-18.522 | 4.735 | (-20.141, -16.904) |
Test
Null hypothesis | H₀: μ₁ - µ₂ = 0 |
Alternative hypothesis | H₁: μ₁ - µ₂ ≠ 0 |
T-Value | DF | P-Value |
-22.64 | 132 | 0.000 |
decision:-
p value = 0.000 <0.05
so, we reject the null hypothesis.
we conclude that,
there is sufficient evidence to claim that there is a difference between weight of apples of farm A and farm B at 0.05 level of significance.
*** if you face any trouble to understand the answer to the problem please mention it in the comment box.if you are satisfied, please give me a LIKE if possible.